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1652. Defuse the Bomb

You have a bomb to defuse, and your time is running out! Your informer will provide you with a circular array code of length of n and a key k.

To decrypt the code, you must replace every number. All the numbers are replaced simultaneously.

  • If k > 0, replace the ith number with the sum of the next k numbers.
  • If k < 0, replace the ith number with the sum of the previous k numbers.
  • If k == 0, replace the ith number with 0.

As code is circular, the next element of code[n-1] is code[0], and the previous element of code[0] is code[n-1].

Given the circular array code and an integer key k, return the decrypted code to defuse the bomb!

Example 1:

Input: code = [5,7,1,4], k = 3
Output: [12,10,16,13]
Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around.

Example 2:

Input: code = [1,2,3,4], k = 0
Output: [0,0,0,0]
Explanation: When k is zero, the numbers are replaced by 0.

Example 3:

Input: code = [2,4,9,3], k = -2
Output: [12,5,6,13]
Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers.

Constraints:

  • n == code.length
  • 1 <= n <= 100
  • 1 <= code[i] <= 100
  • -(n - 1) <= k <= n - 1

Solutions (Ruby)

1. Solution

# @param {Integer[]} code
# @param {Integer} k
# @return {Integer[]}
def decrypt(code, k)
  n = code.length
  s = k > 0 ? 1 : n + k
  e = k > 0 ? k + 1 : n
  sum = code[s...e].sum
  ret = Array.new(n)

  (0...n).each do |i|
    ret[i] = sum
    sum += code[(e + i) % n] - code[(s + i) % n]
  end

  ret
end

Solutions (Rust)

1. Solution

impl Solution {
    pub fn decrypt(code: Vec<i32>, k: i32) -> Vec<i32> {
        let n = code.len();
        let s = if k > 0 { 1 } else { (n as i32 + k) as usize };
        let e = if k > 0 { k as usize + 1 } else { n };
        let mut sum = code[s..e].iter().sum();
        let mut ret = vec![0; n];

        for i in 0..n {
            ret[i] = sum;
            sum += code[(e + i) % n] - code[(s + i) % n];
        }

        ret
    }
}